Trigonometry question Watch

lalyala
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#1
Report Thread starter 7 years ago
#1
The first bit of the question had to do with proving the following using De Moivre's theorem.
sin(3&) = 3 sin & -(4 sin &)^3 (1)
I've done that part fine but I can't seem to do the next part :
Deduce from the above equation that :
sin(pi/12) = (sqrt6 - sqrt2)/4

Using (1) :
sin(pi/12) = 3 sin (pi/36) - (4 sin(pi/36))^3
However sin(pi/36) doesn't come out with a very nice number and I can't seem to figure out which trigonometric identity would be good for this. I feel like its a simple question and I just can't see the solution driving me crazy! Some help would be greatly appreciated!
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fidan134
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Report 7 years ago
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replace (sin&)^3 by sin&*(1-(cos&)^2)
then use cosine double angle formula to express (cos&)^2 by (cos2&+1)/2
Now:
sin(3*pi/12)=3sin(pi/12)-4sin(pi/12)*((cos2&+1)/2)
sin(3*pi/12)=3sin(pi/12)-4sin(pi/12)*((cos2*(pi/12)+1)/2)
sin(3*pi/12)=3sin(pi/12)-4sin(pi/12)*((cos(pi/6)+1)/2)
now your sin(3*pi/12)=root two over two, from calculator
(cos(pi/6)+1)/2=(2+root3)/4 from calculator as well
sub this values and rearrange equation as sin(pi/12) equals to ...
hope that makes sense
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Piecewise
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(Original post by lalyala)
...
Are you given that \sin{\frac{\pi}{4}} = \cos{\frac{\pi}{4}} = \frac{1}{\sqrt{2}}?
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